In my physics textbook, it says that for any pulse, if $\Delta x$ becomes smaller, $\Delta k$ becomes larger where $k$ refers to $2\pi/\lambda$ and $x$ is x-axis displacement, as described by $\Delta x \Delta k \approx 1$. Why is it like this?

2 Answers
2

It's not clear whether you mean a pulse of a wave, i.e. a short section of a wave, or whether you mean a top hat function, but in both cases the principle is the same.

If you Fourier transform a pulse of a wave or a top hat function you get the frequencies that make it up. If you decrease the length of the pulse or reduce the width of the top hat function you'll find that the width of your Fourier transform increases i.e. it spreads across more frequencies. That means it's harder and harder to pin down what you mean by the frequency of the pulse. In the limit of reducing your pulse to a delta function you find the Fourier transform now inludes all frequencies from zero to infinity at an equal amplitude so it's impossible to define even an average frequency.

This is the sense in which $\Delta k$ becomes larger. In real life we often find the Fourier transform is approximately gaussian and we can define $\Delta k$ as the 1/e half width (i.e. the standard deviation).

To add a little to John Rennie's answer--- it is only true for pulses that are smooth and qualitatively resemble Gaussians, for which the root mean square widths of the pulses are related by the exact relation:

$$ \Delta x \Delta k = {1\over 2}$$

If you make a sum of separated delta function pulses, you can make both $\Delta X$ and $\Delta k$ big. For example, taking a signal which is two delta functions at $x=\pm a$, the Fourier transform is a sum of two sinusoids which are nonzero over all space, so $\Delta k$ and $\Delta x$ can both be large.

If you make a non-smooth pulse, you can have large tails on the k-distribution. For example, the Fourier transform of $e^{-|x|}$ pulse is ${1\over k^2+1}$, which has an infinite second moment, so is infinitely broad in k (but not infinitely sharp in x).